How you demonstrate that $C_n^0-C_n^2+C_n^4-C_n^6+...=2^(n/2)cos((npi)/4)$ ?

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Answer 1 · 2017-05-08T22:19:49

See below.

$1/2((1+ i x)^n+(1-i x)^n)=sum_(k=0)^(floor(n/2))(-1)^k((2k),(n))x^(2k)$

or with $x=1$

$1/2((1+ i)^n+(1-i)^n)=sum_(k=0)^(floor(n/2))(-1)^k((2k),(n))$

Now

$1+i = sqrt2 e^(ipi/4)$
$1-i = sqrt2 e^(-ipi/4)$

then

$1/2((1+ i)^n+(1-i)^n)= (sqrt2)^ncos((n pi)/4)$

so finally

$sum_(k=0)^(floor(n/2))(-1)^k((2k),(n)) = 2^(n/2)cos((n pi)/4)$

How you demonstrate that C_n^0-C_n^2+C_n^4-C_n^6+...=2^(n/2)cos((npi)/4)C_n^0-C_n^2+C_n^4-C_n^6+...=2^(n/2)cos((npi)/4)?